Given an array, rearrange the array such that :
- If index i is even, arr[i] <= arr[i+1]
- If index i is odd, arr[i] >= arr[i+1]
Note : There can be multiple answers.
Examples:
Input : arr[] = {2, 3, 4, 5}
Output : arr[] = {2, 4, 3, 5}
Explanation : Elements at even indexes are
smaller and elements at odd indexes are greater
than their next elements
Note : Another valid answer
is arr[] = {3, 4, 2, 5}
Input :arr[] = {6, 4, 2, 1, 8, 3}
Output :arr[] = {4, 6, 1, 8, 2, 3}
This problem is similar to sort an array in wave form.
A simple solution is to sort the array in decreasing order, then starting from second element, swap the adjacent elements.
An efficient solution is to iterate over the array and swap the elements as per the given condition.
If we have an array of length n, then we iterate from index 0 to n-2 and check the given condition.
At any point of time if i is even and arr[i] > arr[i+1], then we swap arr[i] and arr[i+1]. Similarly, if i is odd and
arr[i] < arr[i+1], then we swap arr[i] and arr[i+1].
For the given example:
Before rearrange, arr[] = {2, 3, 4, 5}
Start iterating over the array till index 2 (as n = 4)
First Step:
At i = 0, arr[i] = 2 and arr[i+1] = 3. As i is even and arr[i] < arr[i+1], don't need to swap.
Second step:
At i = 1, arr[i] = 3 and arr[i+1] = 4. As i is odd and arr[i] < arr[i+1], swap them.
Now arr[] = {2, 4, 3, 5}
Third step:
At i = 2, arr[i] = 3 and arr[i+1] = 5. So, don’t need to swap them
After rearrange, arr[] = {2, 4, 3, 5}
C++
// CPP code to rearrange an array such that // even index elements are smaller and odd // index elements are greater than their // next. #include <iostream> using namespace std; void rearrange(int* arr, int n) { for (int i = 0; i < n - 1; i++) { if (i % 2 == 0 && arr[i] > arr[i + 1]) swap(arr[i], arr[i + 1]); if (i % 2 != 0 && arr[i] < arr[i + 1]) swap(arr[i], arr[i + 1]); } } /* Utility that prints out an array in a line */void printArray(int arr[], int size) { for (int i = 0; i < size; i++) cout << arr[i] << " "; cout << endl; } /* Driver function to test above functions */int main() { int arr[] = { 6, 4, 2, 1, 8, 3 }; int n = sizeof(arr) / sizeof(arr[0]); cout << "Before rearranging: \n"; printArray(arr, n); rearrange(arr, n); cout << "After rearranging: \n"; printArray(arr, n); return 0; } |
Java
// Java code to rearrange an array such // that even index elements are smaller // and odd index elements are greater // than their next. class GFG { static void rearrange(int arr[], int n) { int temp; for (int i = 0; i < n - 1; i++) { if (i % 2 == 0 && arr[i] > arr[i + 1]) { temp = arr[i]; arr[i] = arr[i + 1]; arr[i + 1] = temp; } if (i % 2 != 0 && arr[i] < arr[i + 1]) { temp = arr[i]; arr[i] = arr[i + 1]; arr[i + 1] = temp; } } } /* Utility that prints out an array in a line */ static void printArray(int arr[], int size) { for (int i = 0; i < size; i++) System.out.print(arr[i] + " "); System.out.println(); } // Driver code public static void main(String[] args) { int arr[] = { 6, 4, 2, 1, 8, 3 }; int n = arr.length; System.out.print("Before rearranging: \n"); printArray(arr, n); rearrange(arr, n); System.out.print("After rearranging: \n"); printArray(arr, n); } } // This code is contributed by Anant Agarwal. |
Python3
# Python code to rearrange # an array such that # even index elements # are smaller and odd # index elements are # greater than their # next. def rearrange(arr, n): for i in range(n - 1): if (i % 2 == 0 and arr[i] > arr[i + 1]): temp = arr[i] arr[i]= arr[i + 1] arr[i + 1]= temp if (i % 2 != 0 and arr[i] < arr[i + 1]): temp = arr[i] arr[i]= arr[i + 1] arr[i + 1]= temp # Utility that prints out an array in # a line def printArray(arr, size): for i in range(size): print(arr[i], " ", end ="") print() # Driver code arr = [ 6, 4, 2, 1, 8, 3 ] n = len(arr) print("Before rearranging: ") printArray(arr, n) rearrange(arr, n) print("After rearranging:") printArray(arr, n); # This code is contributed # by Anant Agarwal. |
C#
// C# code to rearrange an array such // that even index elements are smaller // and odd index elements are greater // than their next. using System; class GFG { static void rearrange(int[] arr, int n) { int temp; for (int i = 0; i < n - 1; i++) { if (i % 2 == 0 && arr[i] > arr[i + 1]) { temp = arr[i]; arr[i] = arr[i + 1]; arr[i + 1] = temp; } if (i % 2 != 0 && arr[i] < arr[i + 1]) { temp = arr[i]; arr[i] = arr[i + 1]; arr[i + 1] = temp; } } } /* Utility that prints out an array in a line */ static void printArray(int[] arr, int size) { for (int i = 0; i < size; i++) Console.Write(arr[i] + " "); Console.WriteLine(); } // Driver code public static void Main() { int[] arr = { 6, 4, 2, 1, 8, 3 }; int n = arr.Length; Console.WriteLine("Before rearranging: "); printArray(arr, n); rearrange(arr, n); Console.WriteLine("After rearranging: "); printArray(arr, n); } } // This code is contributed by vt_m. |
PHP
<?php // PHP code to rearrange an array such // that even index elements are smaller // and odd index elements are greater // than their next. function swap(&$a, &$b) { $temp = $a; $a = $b; $b = $temp; } function rearrange(&$arr, $n) { for ($i = 0; $i < $n - 1; $i++) { if ($i % 2 == 0 && $arr[$i] > $arr[$i + 1]) swap($arr[$i], $arr[$i + 1]); if ($i % 2 != 0 && $arr[$i] < $arr[$i + 1]) swap($arr[$i], $arr[$i + 1]); } } /* Utility that prints out an array in a line */function printArray(&$arr, $size) { for ($i = 0; $i < $size; $i++) echo $arr[$i] . " "; echo "\n"; } // Driver Code $arr = array(6, 4, 2, 1, 8, 3 ); $n = sizeof($arr); echo "Before rearranging: \n"; printArray($arr, $n); rearrange($arr, $n); echo "After rearranging: \n"; printArray($arr, $n); // This code is contributed // by ChitraNayal ?> |
Output:
Before rearranging: 6 4 2 1 8 3 After rearranging: 4 6 1 8 2 3
Time Complexity :O(n)
Reference :
discuss.codechef.com/questions/43062/anuund-editorial
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